Question

1. Show that: [ x l m 1]
[ a x n 1] = (x-a)(x-b)(x-c)
[ a b x 1]
[ a b c 1]

2. Show that: [ 1+ (a)^2 - (b)^2 2b -2b ]
[ 2ab 1 - (a)^2 + (b)^2 2a ]
[ 2b -2(a)^2 1 - (a)^2 - (b)^2 ]
is a perfect cube.

{please note that these are determinants, the first one is a 4X4 determinant and the second one is a 3X3 determinant}

Answer 1

Answer on Question #46738 – Math – Linear Algebra

Problem.

1. Show that: [ x l m 1 ]

[ a x n 1 ] = (x - a)(x - b)(x - c)

[ a b x 1 ]

[ a b c 1 ]

2. Show that: [ 1 + (a)^2 - (b)^2 2b - 2b ]

[ 2ab 1 - (a)^2 + (b)^2 2a ]

[ 2b - 2(a)^2 1 - (a)^2 - (b)^2 ]

is a perfect cube.

{please note that these are determinants, the first one is a 4X4 determinant and the second one is a 3X3 determinant}

Solution:

1.

$$\left[ \begin{array}{cccc} x & l & m & 1 \\ a & x & n & 1 \\ a & b & x & 1 \\ a & b & c & 1 \end{array} \right] \sim \left[ \begin{array}{cccc} x - a & l - b & m - c & 0 \\ 0 & x - b & n - c & 0 \\ 0 & 0 & x - c & 0 \\ a & b & c & 1 \end{array} \right]$$

Hence

$$\det \left[ \begin{array}{cccc} x - a & l - b & m - c & 0 \\ 0 & x - b & n - c & 0 \\ 0 & 0 & x - c & 0 \\ a & b & c & 1 \end{array} \right] = 1 \cdot \det \left[ \begin{array}{cccc} x - a & l - b & m - c \\ 0 & x - b & n - c \\ 0 & 0 & x - c \end{array} \right] = (x - a)(x - b)(x - c)$$

Then

$$\det \left[ \begin{array}{cccc} x & l & m & 1 \\ a & x & n & 1 \\ a & b & x & 1 \\ a & b & c & 1 \end{array} \right] = (x - a)(x - b)(x - c)$$

2.

$$\det \left[ \begin{array}{ccc} 1 + a^2 - b^2 & 2b & -2b \\ 2ab & 1 - a^2 + b^2 & 2a \\ 2b & -2a^2 & 1 - a^2 - b^2 \end{array} \right]$$

isn't a perfect cube for all $a$ and $b$, as for $a = 2$ and $b = 0$

$$\begin{array}{l} \det \left[ \begin{array}{ccc} 1 + a^2 - b^2 & 2b & -2b \\ 2ab & 1 - a^2 + b^2 & 2a \\ 2b & -2a^2 & 1 - a^2 - b^2 \end{array} \right] \\ = \det \left[ \begin{array}{ccc} 5 & 0 & 0 \\ 0 & -3 & 4 \\ 0 & -8 & -3 \end{array} \right] = 205 \end{array}$$

and 205 isn't a perfect cube.

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